AP® BIOLOGY EQUATIONS AND FORMULAS
Statistical Analysis and Probability
Mean Standard Deviation x = sample mean
n 2
1 Â (xi - x ) n = sample size
nÂ
x = xi s =
i=1 n -1
s = sample standard deviation (i.e., the sample-based
Standard Error of the Mean Chi-Square estimate of the standard deviation of the
population)
SE x =
s o e 2
n 2 e o = observed results
Chi-Square Table e = expected results
p Degrees of Freedom
value = sum of all
1 2 3 4 5 6 7 8
0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51
Degrees of freedom are equal to the number of
0.01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 distinct possible outcomes minus one.
Laws of Probability Metric Prefixes
If A and B are mutually exclusive, then:
Factor Prefix Symbol
P(A or B) = P(A) + P(B)
10 9 giga G
If A and B are independent, then: 10 6 mega M
P(A and B) = P(A) P(B) 10 3 kilo k
10 – 1 deci d
Hardy-Weinberg Equations
10 – 2 centi c
p2 + 2pq + q2 = 1 p = frequency of allele 1 in a
10 – 3 milli m
population
p+q=1 10 – 6 micro μ
q = frequency of allele 2 in a 10 – 9 nano n
population 10 – 12 pico p
Mode = value that occurs most frequently in a data set
Median = middle value that separates the greater and lesser halves of a data set
Mean = sum of all data points divided by number of data points
Range = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum)
8 AP Biology Practice Exam
, Rate and Growth Water Potential ( Y )
Rate dY = amount of change Y = Y P + YS
dY dt = change in time
dt YP = pressure potential
B = birth rate
Population Growth
D = death rate YS = solute potential
dN
= B-D N = population size
dt The water potential will be equal to the
Exponential Growth K = carrying capacity solute potential of a solution in an open
container because the pressure potential of
rmax = maximum per capita
dN the solution in an open container is zero.
= rmax N growth rate of population
dt
The Solute Potential of a Solution
Logistic Growth
( )
YS = -iCRT
dN K-N
= rmax N
dt K i = ionization constant (1.0 for sucrose
because sucrose does not ionize in
Simpson’s Diversity Index water)
( )
2
n C = molar concentration
Diversity Index = 1 - Â
N
R = pressure constant
݊ ൌ total number of organisms of a particular species
( R = 0.0831 liter bars/mole K)
ܰ ൌ total number of organisms of all species T = temperature in Kelvin (ºC + 273)
pH = – log[H+]
Surface Area and Volume
Surface Area of a Sphere Volume of a Sphere r = radius
SA 4 r 2 4
V r3 l = length
3
Surface Area of a Rectangular Volume of a Rectangular Solid h = height
Solid V lwh
w = width
SA 2lh 2lw 2wh
Volume of a Cylinder s = length of one
Surface Area of a Cylinder V r 2h side of a
SA 2 rh 2 r 2 cube
Volume of a Cube SA = surface area
Surface Area of a Cube V s3
SA 6s 2 V = volume
AP Biology Practice Exam 9
, BIOLOGY
SECTION I
Time—1 hour and 30 minutes
60 Questions
Directions: Each of the questions or incomplete statements below is followed by four suggested answers or
completions. Select the one that is best in each case and then enter the letter in the corresponding space on the
answer sheet.
1. Many species of corals are threatened by the 2. A researcher measured the temperature at which
increasing temperatures and decreasing pH of two different samples of double-stranded DNA
ocean waters. One species, denature (separate into single strands). Sample
Stylophora pistillata, has been found to thrive in 1 denatured at a significantly lower temperature
water that is warmer and has a lower pH than than sample 2 did. Based on the data, the
the water that corals typically thrive in. researcher claims that the DNA in sample 2 is
Additionally, researchers have found that the composed of a higher percentage of guanine and
tolerance for the new water conditions is cytosine than the DNA in sample 1 is.
heritable.
Which of the following best supports the
Which of the following statements best explains researcher’s claim?
the changes seen in S. pistillata in response to
the changing water conditions? (A) The bonds between guanine and cytosine
are covalent bonds, which require more
(A) The corals’ adaptation is an example of energy to disrupt than those between
natural selection because the tolerance is adenine and thymine.
in response to a changing environment (B) Guanine-cytosine pairs denature at a higher
and has a genetic basis. temperature because they have more
(B) The corals’ adaptation is an example of the hydrogen bonds between them than
founder effect because the majority of adenine-thymine pairs do.
corals do not have a tolerance for warmer (C) Adenine-thymine pairs require less energy
water. to separate because adenine and thymine
(C) The corals’ adaptation is an example of are both single-ring bases.
genetic drift because the change was a (D) Guanine-cytosine pairs require more
chance event and not the result of energy to separate because one is a purine
selection in response to environmental and one is a pyrimidine.
change.
(D) The corals’ adaptation is an example of
adaptive radiation because it has resulted
in a wide range of species adapting to the
new ocean conditions.
GO ON TO THE NEXT PAGE.
10 AP Biology Practice Exam
, 3. Researchers studying the bacterium Escherichia coli split a population of the bacteria into two samples.
Sample 1 was transformed with a plasmid containing a gene that makes the bacteria resistant to the antibiotic
kanamycin. Sample 2 was transformed with a plasmid lacking the antibiotic resistance gene. A portion of
each sample was then added to growth plates containing just nutrients or growth plates containing nutrients and
kanamycin.
After being allowed to grow for 24 hours at 37˜C , the number of colonies on each growth plate was counted
(Table 1).
TABLE 1. BACTERIAL GROWTH FOLLOWING TRANSFORMATION WITH OR
WITHOUT THE KANAMYCIN RESISTANCE GENE
Transforming Plasmid
Plate # Contains Kanamycin Growth Media Results
Resistance Gene
1 Too many colonies to
No Nutrient only
count (lawn)
2 Too many colonies to
Yes Nutrient only
count (lawn)
3 No Nutrient with antibiotic No colonies observed
4 Yes Nutrient with antibiotic 7 colonies observed
Which of the following claims is best supported by the data in Table 1 ?
(A) The transformation procedure killed all the bacteria that were added to plate 3.
(B) More bacteria on plates 1 and 2 were successfully transformed than on any other plate.
(C) None of the bacteria on plate 2 were successfully transformed with the kanamycin resistance gene.
(D) Only the bacteria that were successfully transformed with the kanamycin resistance gene grew on
plate 4 .
GO ON TO THE NEXT PAGE.
AP Biology Practice Exam 11
Statistical Analysis and Probability
Mean Standard Deviation x = sample mean
n 2
1 Â (xi - x ) n = sample size
nÂ
x = xi s =
i=1 n -1
s = sample standard deviation (i.e., the sample-based
Standard Error of the Mean Chi-Square estimate of the standard deviation of the
population)
SE x =
s o e 2
n 2 e o = observed results
Chi-Square Table e = expected results
p Degrees of Freedom
value = sum of all
1 2 3 4 5 6 7 8
0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51
Degrees of freedom are equal to the number of
0.01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 distinct possible outcomes minus one.
Laws of Probability Metric Prefixes
If A and B are mutually exclusive, then:
Factor Prefix Symbol
P(A or B) = P(A) + P(B)
10 9 giga G
If A and B are independent, then: 10 6 mega M
P(A and B) = P(A) P(B) 10 3 kilo k
10 – 1 deci d
Hardy-Weinberg Equations
10 – 2 centi c
p2 + 2pq + q2 = 1 p = frequency of allele 1 in a
10 – 3 milli m
population
p+q=1 10 – 6 micro μ
q = frequency of allele 2 in a 10 – 9 nano n
population 10 – 12 pico p
Mode = value that occurs most frequently in a data set
Median = middle value that separates the greater and lesser halves of a data set
Mean = sum of all data points divided by number of data points
Range = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum)
8 AP Biology Practice Exam
, Rate and Growth Water Potential ( Y )
Rate dY = amount of change Y = Y P + YS
dY dt = change in time
dt YP = pressure potential
B = birth rate
Population Growth
D = death rate YS = solute potential
dN
= B-D N = population size
dt The water potential will be equal to the
Exponential Growth K = carrying capacity solute potential of a solution in an open
container because the pressure potential of
rmax = maximum per capita
dN the solution in an open container is zero.
= rmax N growth rate of population
dt
The Solute Potential of a Solution
Logistic Growth
( )
YS = -iCRT
dN K-N
= rmax N
dt K i = ionization constant (1.0 for sucrose
because sucrose does not ionize in
Simpson’s Diversity Index water)
( )
2
n C = molar concentration
Diversity Index = 1 - Â
N
R = pressure constant
݊ ൌ total number of organisms of a particular species
( R = 0.0831 liter bars/mole K)
ܰ ൌ total number of organisms of all species T = temperature in Kelvin (ºC + 273)
pH = – log[H+]
Surface Area and Volume
Surface Area of a Sphere Volume of a Sphere r = radius
SA 4 r 2 4
V r3 l = length
3
Surface Area of a Rectangular Volume of a Rectangular Solid h = height
Solid V lwh
w = width
SA 2lh 2lw 2wh
Volume of a Cylinder s = length of one
Surface Area of a Cylinder V r 2h side of a
SA 2 rh 2 r 2 cube
Volume of a Cube SA = surface area
Surface Area of a Cube V s3
SA 6s 2 V = volume
AP Biology Practice Exam 9
, BIOLOGY
SECTION I
Time—1 hour and 30 minutes
60 Questions
Directions: Each of the questions or incomplete statements below is followed by four suggested answers or
completions. Select the one that is best in each case and then enter the letter in the corresponding space on the
answer sheet.
1. Many species of corals are threatened by the 2. A researcher measured the temperature at which
increasing temperatures and decreasing pH of two different samples of double-stranded DNA
ocean waters. One species, denature (separate into single strands). Sample
Stylophora pistillata, has been found to thrive in 1 denatured at a significantly lower temperature
water that is warmer and has a lower pH than than sample 2 did. Based on the data, the
the water that corals typically thrive in. researcher claims that the DNA in sample 2 is
Additionally, researchers have found that the composed of a higher percentage of guanine and
tolerance for the new water conditions is cytosine than the DNA in sample 1 is.
heritable.
Which of the following best supports the
Which of the following statements best explains researcher’s claim?
the changes seen in S. pistillata in response to
the changing water conditions? (A) The bonds between guanine and cytosine
are covalent bonds, which require more
(A) The corals’ adaptation is an example of energy to disrupt than those between
natural selection because the tolerance is adenine and thymine.
in response to a changing environment (B) Guanine-cytosine pairs denature at a higher
and has a genetic basis. temperature because they have more
(B) The corals’ adaptation is an example of the hydrogen bonds between them than
founder effect because the majority of adenine-thymine pairs do.
corals do not have a tolerance for warmer (C) Adenine-thymine pairs require less energy
water. to separate because adenine and thymine
(C) The corals’ adaptation is an example of are both single-ring bases.
genetic drift because the change was a (D) Guanine-cytosine pairs require more
chance event and not the result of energy to separate because one is a purine
selection in response to environmental and one is a pyrimidine.
change.
(D) The corals’ adaptation is an example of
adaptive radiation because it has resulted
in a wide range of species adapting to the
new ocean conditions.
GO ON TO THE NEXT PAGE.
10 AP Biology Practice Exam
, 3. Researchers studying the bacterium Escherichia coli split a population of the bacteria into two samples.
Sample 1 was transformed with a plasmid containing a gene that makes the bacteria resistant to the antibiotic
kanamycin. Sample 2 was transformed with a plasmid lacking the antibiotic resistance gene. A portion of
each sample was then added to growth plates containing just nutrients or growth plates containing nutrients and
kanamycin.
After being allowed to grow for 24 hours at 37˜C , the number of colonies on each growth plate was counted
(Table 1).
TABLE 1. BACTERIAL GROWTH FOLLOWING TRANSFORMATION WITH OR
WITHOUT THE KANAMYCIN RESISTANCE GENE
Transforming Plasmid
Plate # Contains Kanamycin Growth Media Results
Resistance Gene
1 Too many colonies to
No Nutrient only
count (lawn)
2 Too many colonies to
Yes Nutrient only
count (lawn)
3 No Nutrient with antibiotic No colonies observed
4 Yes Nutrient with antibiotic 7 colonies observed
Which of the following claims is best supported by the data in Table 1 ?
(A) The transformation procedure killed all the bacteria that were added to plate 3.
(B) More bacteria on plates 1 and 2 were successfully transformed than on any other plate.
(C) None of the bacteria on plate 2 were successfully transformed with the kanamycin resistance gene.
(D) Only the bacteria that were successfully transformed with the kanamycin resistance gene grew on
plate 4 .
GO ON TO THE NEXT PAGE.
AP Biology Practice Exam 11
